Question

The compound 1-iodododecane is a nonvolatile liquid with a density of 1.20g/mL. The density of mercury is 13.6g/mL. What do you predict for the height of a barometer column based on 1-iodododecane, when the atmospheric pressure is 749 torr? What is the pressure, in atmospheres, on the body of a diver if he is 21 ft below the surface of the water when the atmospheric pressure is 742 torr?

Answer 1

a.
`h=0.823m`

b.
`P_(diver)=161653.04Pa`

Step-by-step explanation:

Hello,

a.

One could consider the following equation for this exercise's first part:

`P_(liq)=P_(h)+P_(atm)\\d_(liq)hg=d_(Hg)hg+P_(atm)\\`

Whereas
`d` is the density,
`g` the acceleration of gravity,
`h` the height and
`P_(atm)` the atmospheric pressure, thus:

`h=(P_(atm))/(g*(d_(liq)-d_(Hg))) =(749torr*(101325Pa)/(760torr) )/(-9.8(m)/(s^2)*(1.2-13.6)(g)/(mL)*(1kg)/(1000g)*(1x10^6mL)/(1m^3) ) \\h=0.823m`

b.

Now, to know the pressure on the diver's body, one uses the following equation considering that the pressure is exerted downwards:

`P_(diver)=P_(atm)+d_(water)gh\\\\P_(diver)=742torr*(101325Pa)/(760torr)+1000(kg)/(m^3) *9.8(m)/(s^2) *21ft*(0.3048m)/(1 ft) \\P_(diver)=161653.04Pa`

Best regards.